"Max Power" Calculations in Electrical Engineering

[Studiot]

This is where I inadvertanly reversed the term order, relative to the first equation.

R5 = 50 so power in R5 is

V²/50

R5 is the only resistor that experiences the full voltage, V.
All the others see V reduced by some factor.
However you still need to check that R5 reaches 1 watt first, since we also divide by the resistor value to get the power.

I am off now for some parvus et circum in the shape of dinner and Alien v Predator 2.

go well

 

[Femme_physics]

I see. It's also a bit late here. I'll post tomorrow the numbers hopefully I'd get someone's seal of approval. Thank you Studiot, Serena.

 

[Femme_physics]

1) I've scanned my second attempt of the solution

2) I've also tried to calculate for I, is my equation correct?

Also, as far as how you've gotten the the formula for each of the resistor, it's basically:

"the potential difference of the resistor (or the entire parallel region), divided by the potential difference across the entire closed loop, divided by the potential difference of the resistor."

I've tried to derive that from your formula. I don't recall it being explained to me that way.

 

[Studiot]

Oh dear, you really are better at mechanics.

I will post some more after lunch. Meanwhile perhaps you would like to think about this.

Why does your 'solution' equate the total power to 1 and then 300?

The current method is not really derivable from the voltage method - V does not actually appear anywhere.

Kirchoffs laws, loops and such won't help here. I'm just using the basic definition of series and parallel connections, ohm's law and the equation for power.

 

[Femme_physics]

Oh dear, you really are better at mechanics.

LOL. Appreciate the recognition though.

I do know how to find Rt and easily break down a parallel+series circuit to an only series circuit, which I initially thought was challenging but I understand the simple principles of it.

Why does your 'solution' equate the total power to 1 and then 300?

Well, I solved for V, so since I had a fraction and 300 was my common denominator, I multiplied both sides...

So quote is not true? ->

"the potential difference of the resistor (or the entire parallel region), divided by the potential difference across the entire closed loop, divided by the potential difference of the resistor."

It seems to be true to how you've decided the formula for each R in your P equation.

 

[Studiot]

There are two separate issues here.

One is arithmetic and I will sort that first by finishing the solution (in the attachment).

Note that in order to back substitute and calculate the total power I only need to calculate V².
I don't need to take the square root and find V.

I have lined up the calculations so the individual contributions from each resistor is shown term by term.

Is there anything you don't follow?

 

[I like Serena]

1) I've scanned my second attempt of the solution

I'll leave the electrical part of the discussion to Studiot, because I think he's doing a good job at it, and he's obviously taking the time and energy to try to explain it (I think he's improving on that too ).

So I'd like to draw your attention to an arithmetic part (which he ignored for now).

In your scan you write one of the terms (2nd line) as:

10 \cdot \left(\frac {30V} {40}\right)^2

Then in the 3rd line you have turned this into:

10 \cdot \left(\frac {30V^2} {40^2}\right)

And in the 4th line this is:

\left(\frac {300V^2} {1600}\right)

This is not quite right. ;)

Can you see what this should be?

 

[Femme_physics]

Great catch, Serena. You two are really too kind. I appreciate the effort. I also re-read the walking through analysis in page 2 just to make sure it's all clear.

This is not quite right. ;)

I'm plugging it into my calc and I now see it doesn't turn up quite right. I seem to have broken a simple rule of parathesis BEFORE exponents.

Apparently, what's right is what Studiot did, but I'm not able to follow him. Did he just use the properties of exponents?

Not sure how he got from this to that

http://img717.imageshack.us/img717/992/howhow.jpg

He seems to have used the exponent on both the numerical denominator and numerator in the fraction and then just pull V out and leave it to the power of 2. I wasn't aware you can do that. He also took the 30 on the lowest denominator and multiplied it by the upper fraction. But at this page I already seemed to have lost Studiot on too many fronts.

Is this all confusion due to my lack of understanding about simplifying exponential fractions containing unknowns? If it is, I'll go back and review it.

 

[I like Serena]

Great catch, Serena. You two are really too kind. I appreciate the effort. I also re-read the walking through analysis in page 2 just to make sure it's all clear.

I'm plugging it into my calc and I now see it doesn't turn up quite right. I seem to have broken a simple rule of parathesis BEFORE exponents.

Apparently, what's right is what Studiot did, but I'm not able to follow him. Did he just use the properties of exponents?

Not sure how he got from this to that

http://img717.imageshack.us/img717/992/howhow.jpg

He seems to have used the exponent on both the numerical denominator and numerator in the fraction and then just pull V out and leave it to the power of 2. I wasn't aware you can do that. He also took the 30 on the lowest denominator and multiplied it by the upper fraction. But at this page I already seemed to have lost Studiot on too many fronts.

Is this all confusion due to my lack of understanding about simplifying exponential fractions containing unknowns? If it is, I'll go back and review it.

I'm focusing on arithmetic now, because you will definitely need that too with calculus and dynamics.

Studiot skipped a few steps, so I'll write them out for you for this specific instance.

He used the following rules for exponents and fractions:

(2 \cdot 3)^2 = 2^2 \cdot 3^2

\left(\frac 2 3\right)^2 = \frac {2^2} {3^2}

3 \cdot \frac 2 3 = 2

\frac {\left( \frac 2 3 \right)} 5 =\frac {3 \cdot \frac 2 3} {3 \cdot 5} = \frac 2 {3 \cdot 5}

\frac {2 \cdot 3} 5=\frac 2 5 \cdot 3


So taking you step by step, applying the rules above, we have:

\frac {\left(\frac {30} {40}V\right)^2} {30}<br /> =\frac {\left(\frac {30} {40} \cdot V\right)^2} {30}<br /> =\frac {\left(\frac {30} {40}\right)^2 \cdot V^2} {30}<br /> = \frac {\frac {30^2} {40^2} \cdot V^2} {30}<br /> = \frac {40^2 \cdot \frac {30^2} {40^2} \cdot V^2} {40^2 \cdot 30}<br /> = \frac {30^2 \cdot V^2} {40^2 \cdot 30}<br /> = \frac {30^2} {40^2 \cdot 30} \cdot V^2<br /> = \left( \frac {30 \cdot 30} {40 \cdot 40 \cdot 30} \right) \cdot V^2

If you do this a lot, you can do it all at once, or otherwise when you write it down, you leave it to the reader to figure out the intermediary steps.

 

[Studiot]

Well I'm glad I didn't have to type all that out.

Thank you ILS!

 

[Femme_physics]

He's incredible! :)

I did, however, try to solve it on my own using the properties of exponent you have written out to me, I Like Serena. I swear, I didn't even look at your final big solution! (attached is my draft attempt and success [there was a moment of desperation which might show in the page]... )

I'll look back at some of the earlier posts now, thanks :)

 

[Femme_physics]

Okay, I am looking back at Studiot post. You've found Ptotal, they're asking for Pmax. They didn't ask "which resistor will be damaged first", they asked "what is the max power you can spread around the circuit without causing damage to any of the resistors"..shouldn't the answer be in terms of Pmax = ?

 

[Studiot]

Yes I should have said P_totalmax not just P_total when I put in the value for V²

The maximum safe power in that network is 2.25 Watts.

At that power input R5 is at its rated limit and if you look carefully R4 is pretty close as well.

Are we all agreed now?

 

[I like Serena]

He's incredible! :)

I did, however, try to solve it on my own using the properties of exponent you have written out to me, I Like Serena. I swear, I didn't even look at your final big solution! (attached is my draft attempt and success [there was a moment of desperation which might show in the page]... )

I'll look back at some of the earlier posts now, thanks :)

I'm proud of you! Your calculation is entirely correct

(However, Studiot had 0.01875 V². How could that be?)

 

[Femme_physics]

I'm proud of you! Your calculation is entirely correct

:) Great.

I always make sure to plug a number to see it makes sense!

Yes I should have said P_totalmax not just P_total when I put in the value for V²

The maximum safe power in that network is 2.25 Watts.

At that power input R5 is at its rated limit and if you look carefully R4 is pretty close as well.

Are we all agreed now?

If you can confirm my own verbal-only walk through analysis

In order to find Pmax we first write up the circuit's formula for total power, by doing that we are able to easily detect which resistor will have the greater reaction to upping of the power dosage, i.e. which resistor will burn first.


So after discovering the culprit, we set it to be LESS THAN or EQUAL to 1. i.e. we use the formula for power again, just this time looking at an individual naughty resistor and setting it equal or less than 1.

That allows us to determine MAX voltage [to the power of 2].

(hmm, interesting. What if the voltage was given to us?)

Having found Vmax^2, we just plug it into the Ptotalmax formula and solve. Voila!

(However, Studiot had 0.01875 V2. How could that be?)

*glances at Studiot...whistles, averts gaze*

 

[I like Serena]

*glances at Studiot...whistles, averts gaze*

I'm afraid you were both right in your calculations.
Can you spot the difference and how it matters?

 

[Studiot]

*glances at Studiot...whistles, averts gaze*

I don't seem to get the joke here.

FF, in post#43 you posted some working where you started with a fraction in brackets and squared it.

Yes your arithmetic in that post was correct.

Unfortunately you had not copied down the fraction correctly so the bottom 30 appeared inside your bracket, but was outside in my original.

??

 

[Femme_physics]

Oh, yea, I put the paranthesis all over, whereas the parenthesis just should've been over the upper fraction, so that's what raised to the power of 2, not the lower part

Edit: I wrote it before Studiot posted!

 

[Femme_physics]

I don't seem to get the joke here.

FF, in post#43 you posted some working where you started with a fraction in brackets and squared it.

Yes your arithmetic in that post was correct.

Unfortunately you had not copied down the fraction correctly so the bottom 30 appeared inside your bracket, but was outside in my original.

??

Yes, my bad, it's just that I have a thing with experts and solution manuals writing the wrong figures so I keep looking for it even when it's not there. Anyway, you're a world of help! :) Can we go back to -->
"
If you can confirm my own verbal-only walk through analysis

In order to find Pmax we first write up the circuit's formula for total power, by doing that we are able to easily detect which resistor will have the greater reaction to upping of the power dosage, i.e. which resistor will burn first.


So after discovering the culprit, we set it to be LESS THAN or EQUAL to 1. i.e. we use the formula for power again, just this time looking at an individual naughty resistor and setting it equal or less than 1.

That allows us to determine MAX voltage [to the power of 2].

(hmm, interesting. What if the voltage was given to us?)

Having found Vmax^2, we just plug it into the Ptotalmax formula and solve. Voila!
"

 

[I like Serena]

Oh, yea, I put the paranthesis all over, whereas the parenthesis just should've been over the upper fraction, so that's what raised to the power of 2, not the lower part

Edit: I wrote it before Studiot posted!

@Studiot: Sorry, I just thought it funny that FP thought you had it wrong, but of course you were right all along.

I did hope that FP would have found that out for herself ;).

 

[Studiot]

I really hate to miss a joke, even one at my expense.

Folks can call me anything they like, just so long as it is clearly in jest.
Laughter is almost as good as a good dinner.

 

[Studiot]

(hmm, interesting. What if the voltage was given to us?)

Surely if the voltage is fixed the power is fixed?

The rest of your summary is good.



How do you feel about the circuit analysis now?

 

[Femme_physics]

Surely if the voltage is fixed the power is fixed?

I get it, so the max voltage is always the max power! Brilliant. Absolutely brillant.

Somebody tell Homer Simpson that! :)

The rest of your summary is good.



How do you feel about the circuit analysis now?

Not...that bad frankly. It's kinda lovely, and makes sense. It's, again, one of those subjects you must love to succeed in it. I love :) I hope to get as good as mechanics in it. I'm glad to have great mentors like you two (if you don't mind me calling you that :) ), that don't even mind going back to the basic groundworks to get up the scale back to electronics.

By the way, do you know how much self-restraint it took not to answer your question with "electrifying"?? lol

Oh, btw, as long as you and ILS are here, I have another coursework question that I... *see them both bolt for their lives* ;)

 

[I like Serena]

*bolting*

--I like ILSe

[wikipedia: "249 Ilse - an asteroid named for the legendary German princess of the Harz Mountains"]

 

[Studiot]

If you are comfortable with basic circuit analysis, you might like to look at this.

https://www.physicsforums.com/showthread.php?t=493068

 

[Femme_physics]

If you are comfortable with basic circuit analysis, you might like to look at this.

https://www.physicsforums.com/showthread.php?t=493068

I was going to solve his problem for him just for practice but even the problem bolted and ran away from me. Don't believe me? Here, look!

 

[I like Serena]

I was going to solve his problem for him just for practice but even the problem bolted and ran away from me. Don't believe me? Here, look!

 

[I like Serena]

I was going to solve his problem for him just for practice but even the problem bolted and ran away from me. Don't believe me? Here, look!

Seriously.
You can't treat R1 and R2 as parallel resistors, since there is an R3 in between.*

However, if you deform the circuit a bit and bring the end points of R5 and R6 together to one point, which you can do because they are short circuited, it becomes a problem I think you can already solve.

In the new deformed circuit the resistors R5 and R6, for instance, are parallel.

----
* Note that I followed your numbering of the resistors, and not the numbering of the thread.

 

[Femme_physics]

Seeing how you can deform the circuits is indeed crucial in solving these types of problems, I was rather shortsighted this time. Thanks, I'll probably won't post in trying to help him or anyone with electronics problem anytime soon.

 

[I like Serena]

Seeing how you can deform the circuits is indeed crucial in solving these types of problems, I was rather shortsighted this time. Thanks, I'll probably won't post in trying to help him or anyone with electronics problem anytime soon.

If by "anytime soon" you mean "after a couple of days", I can agree. :)

And anyway, I'm pretty sure any contribution you make would be appreciated.
It shows how enthusiastic you are, and you already know how well people like that and respond to that.
Imho ppl respond much less to a "know it all" professor, even if he/she is right.

(Just try not to hijack any such thread )